Chapter 5 Notes
 The atomic mass of an element is a
weighted average mass of the atoms
found in nature.
 If you were to mass an oxygen atom,
would it weigh 15.9994 amu?
 NO! 15.999 amu – THIS IS ONLY AN
AVERAGE OF ALL THE ISOTOPES OF
OXYGEN 16O, 17O, 18O
The relative abundance of
each isotope of an element
determines its atomic mass.
Naturally occurring
isotope
Fractional
abundance
Mass (amu)
11C
0
11.011430
12C
0.989
12.000000
13C
0.011
13.003355
14C
0
14.003241
 Three isotopes of oxygen occur in
nature: oxygen-16, oxygen-17, and
oxygen-18. The atomic mass of
oxygen is 15.9994 because oxygen-16
is the most abundant.
QUESTION
 If element Z has two naturally
occurring isotopes: Z-20 and Z-22 and
the atomic mass of Z is 21.5 amu
 Which isotope occurs most often in
nature?
 Z-22
 the relative abundance of each isotope
in nature is one Z-20 to two Z-22.
 Thallium has two isotopes, thallium-203 and
thallium-205. Thallium’s atomic number is
81 and its atomic mass is 204.38 amu.
 So, is there more Thallium-203 or Thallium205?
 There is more Thallium-205 found in nature.
 Atoms of both isotopes have 81 protons.
Thallium-205 atoms have more neutrons.
 A sample of oxygen contains three
naturally occurring isotopes:
16
8
O
17
8
O
18
8
O
The relative abundances
and atomic masses are:
16
8
O
 99.7759% (mass = 15.995 amu)
17
8
18
8
O
O
 0.037% (mass = 16.995 amu)
 0.204% (mass = 17.999 amu)
 Calculate the average atomic mass of the oxygen.
 (percent x mass) + (percent x mass) and
so on….
 0.99759(15.99491) +
 0.00037(16.99913) +
 0.00204(17.99916) =
 15.999 amu
Radioactivity (Ch 25)
 An alpha particle (radiation) is a
helium nucleus. 4
2
U 
238
92
He
Th  He
234
90
4
2
 A beta particle (radiation) is an
electron. 0
e
1
14
6
C 
N e
14
7
0
1
Gamma Radiation
 Gamma radiation is high-frequency
electromagnetic radiation. Gamma
radiation has no mass or charge so
there is no change in the atomic
number caused by the emission of
gamma radiation.
 Nuclear equations are balanced.
Complete these nuclear
equations.
3
1
H 
 e
0
1

28
13
Al  e
0
1
231
91
Pa 
227
89
Ac 
75
34
Se 
 e
0
1

237
93
Np  He
4
2